# Finding the distribution of $Y = aX + b$ given that $X \sim N(\mu = \alpha, \sigma^{2} = \beta)$.

How do I find the distribution of $$Y = aX + b$$ given that $$X \sim N(\mu = \alpha, \sigma^{2} = \beta)$$.

Let's say for example I have $$X \sim N(\mu = 3, \sigma^{2} = 16)$$ and want to find out the distribution of $$Y = 2X - 5$$ and compute some probability $$P(2X-5 > 0)$$, how would I go about doing this?

Does multiplying a r.v $$X$$ with a constant $$a$$ and then adding a constant $$b$$ change the distribution or will the new r.v $$Y = 2X-5$$ still be normally distributed with the same mean $$\alpha$$ and variance $$\beta$$ as $$X$$?

$$Y$$ still has the normal distribution since it is a linear transformation of a normally distributed random variable $$X$$. However, the mean and the variance are not the same. We need to calculate the expected value and the variance to determine the parameters of the distribution of $$Y$$. We have that $$\operatorname E[Y]=\operatorname E[aX+b]=a\operatorname E[X]+b=a\alpha+b$$ and $$\operatorname{Var}[Y]=\operatorname{Var}[aX+b]=a^2\operatorname{Var}[X]=a^2\beta.$$ Hence, $$Y\sim N(a\alpha+b,a^2\beta)$$.

This holds for normally distributed random variables but does not necessarily hold for other distributions (for example, it does not hold for the Bernoulli distribution).

• A minor point of clarification. The fact that $Y$ remains normal under such a transformation is a property of $X$ having a normal distribution, but the mean and variance being $a \alpha + b$ and $a^2 \beta$, respectively, is true even when $X$ is not normally distributed. I just did not want the OP to think that those formulas only apply when $X$ is normal. Commented Oct 19, 2021 at 18:42

Alternatively, you can use the X variable moment-generating function since $$X \sim N(\alpha, \beta)$$. You only have to check $$Y$$ has a moment-generating function similar to a Normal distribution's one.

By definition $$m_Y(t) = E[e^{tY}]$$ $$= E[e^{t(aX+b)}]$$ $$= E[e^{taX}e^{tb}]$$ $$= e^{tb}E[e^{taX}]$$ $$= e^{tb}(m_X(at))$$ $$= e^{tb}(e^{\alpha(at) + \frac{\beta (at)^2}{2}})$$ $$= e^{(\alpha a) t + tb + \frac{\beta (at)^2}{2}}$$ $$= e^{(\alpha a + b)t + \frac{(\beta a^2) t^2}{2}}$$

Let $$\mu =(\alpha a + b)$$ and $$\sigma = (\beta a^2)$$. Then $$m_Y(t) = e^{\mu t + \frac{\sigma t^2}{2}}$$

As you may see this is a Normal distribution moment-generating function. By inspection, you can check its parameters are as follows: $$E[Y] = (\alpha a + b)$$ and $$Var[Y] = \beta a^2$$.

Then $$Y \sim N(\alpha a + b, \beta a^2)$$.